It can be deduced that the pH of the solution will be 1.42.
The following can be deduced from the information
Mass = 0.73 g of HCl
Volume = 500 ml = 0.5l
Molecular weight of HCl = 36.5 g
From the given information, the mole will be represented by x and is calculated thus:
x = (0.73 x 1) / 36.5
x = 0.019 mol
Then, we'll calculate the molarity which will be:
Molarity = 0.019 / 0.5
Molarity = 0.038
Then, the pH will be calculated thus:
pH = -log [H+]
pH = - log [0.038]
pH = 1.42
In conclusion, the pH is 1.42.
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